A158565 A modulo two based Pascal's triangle using powers of two for even and powers of three for odd: t(n,m)=If[Mod[Binomial[n, m], 2] == 0 && m <= Floor[n/2], 2^m, If[Mod[Binomial[n, m], 2] == 0 && m > Floor[n/2], 2^(n - m), If[Mod[Binomial[n, m], 2] == 1 && m <= Floor[n/2], 3^m, If[Mod[Binomial[n, m], 2] == 1 && m > Floor[n/2], 3^(n - m), 0]]]].

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 2, 4, 2, 1, 1, 3, 4, 4, 3, 1, 1, 2, 9, 8, 9, 2, 1, 1, 3, 9, 27, 27, 9, 3, 1, 1, 2, 4, 8, 16, 8, 4, 2, 1, 1, 3, 4, 8, 16, 16, 8, 4, 3, 1, 1, 2, 9, 8, 16, 32, 16, 8, 9, 2, 1

Offset

0,5

Comments

Row sums are:

{1, 2, 4, 8, 10, 16, 32, 80, 46, 64, 104,...}.

Links

Table of n, a(n) for n=0..65.

Formula

t(n,m)=If[Mod[Binomial[n, m], 2] == 0 && m <= Floor[n/2], 2^m,

If[Mod[Binomial[n, m], 2] == 0 && m > Floor[n/2], 2^(n - m),

If[Mod[Binomial[n, m], 2] == 1 && m <= Floor[n/2], 3^m,

If[Mod[Binomial[n, m], 2] == 1 && m > Floor[n/2], 3^(n - m), 0]]]].

Example

{1},
{1, 1},
{1, 2, 1},
{1, 3, 3, 1},
{1, 2, 4, 2, 1},
{1, 3, 4, 4, 3, 1},
{1, 2, 9, 8, 9, 2, 1},
{1, 3, 9, 27, 27, 9, 3, 1},
{1, 2, 4, 8, 16, 8, 4, 2, 1},
{1, 3, 4, 8, 16, 16, 8, 4, 3, 1},
{1, 2, 9, 8, 16, 32, 16, 8, 9, 2, 1}

Mathematica

Table[Table[If[Mod[Binomial[n, m], 2] == 0 && m <= Floor[n/2], 2^m,
If[Mod[Binomial[n, m], 2] == 0 && m > Floor[n/2], 2^(n - m),
If[Mod[Binomial[n, m], 2] == 1 && m <= Floor[n/2], 3^m,
If[Mod[Binomial[n, m], 2] == 1 && m > Floor[n/2], 3^(n - m),
0]]]], {m, 0, n}], {n, 0, 10}];
Flatten[%]

Crossrefs

Sequence in context: A093557 A098802 A048804 * A132422 A065133 A343033

Adjacent sequences: A158562 A158563 A158564 * A158566 A158567 A158568

Keyword

nonn,tabl,uned

Author

Roger L. Bagula, Mar 21 2009

Status

approved